A thin uniform rod of mass M is suspended horizontally bytwo vertical wires. One

Question

A thin uniform rod of mass M is suspended horizontally bytwo vertical wires. One wire is at the left end of the rod, and theother wire is 3/4 of the length of therod from the left end. (a) Determine the tension (T) in eachwire. (Use g for the acceleration due to gravity and M asnecessary.) TL = TR =
(b) An object is now hung by a string attached to the right end ofthe rod. When this happens, it is noticed that the rod remainshorizontal but the tension in the wire on the left vanishes.Determine the mass m of the object. (Use M for the mass ofthe rod as necessary.)

(a) Determine the tension (T) in eachwire. (Use g for the acceleration due to gravity and M asnecessary.) TL = TR =
(b) An object is now hung by a string attached to the right end ofthe rod. When this happens, it is noticed that the rod remainshorizontal but the tension in the wire on the left vanishes.Determine the mass m of the object. (Use M for the mass ofthe rod as necessary.)
TL = TR =

Explanation / Answer

(a) Determine the tension (T)in each wire. (Use g for the acceleration due to gravity and M asnecessary.) First we know that the vertical forces must sum to 0 an dwehave TL + TR= Mg Moments about any point must also sum to 0. Let's take momentabout the point where right wire is attached (point atTR ) using convention CCW isnegative -(3/4)LTL + 1/4LMg=0 where L - lengthof the rod Now we have to solve for TL and then forTR . TL = (1/3)Mg and TR= Mg - TL TR= Mg - (1/3)Mg TR= (2/3)Mg
(b) An object is now hung by a string attached to the right end ofthe rod. When this happens, it is noticed that the rod remainshorizontal but the tension in the wire on the left vanishes.Determine the mass m of the object. (Use M for the mass ofthe rod as necessary.)
Hmmm..."the tension in the wire on the leftvanishes". Could taht be magic? It can only vanish is the center ofmass of the rod had suddenly shifted to the right and is under thestring or the left side has received a support. 1. In case thecenter of mass has shifted to the right then the tension inthe string is Ts Ts=Mg 2. In case theleft side has received a support then the tension in the string isTs Ts +R=Mg and again usingthe moment say about the far left LTs - 0.5LMg=0 Ts=0.5Mg

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